If y = (x log x)^{log log x},then frac{dy}{dx | vedclass

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(A) Given $y = (x \log x)^{\log \log x}$.Taking logarithm on both sides: $\log y = \log \log x \cdot \log(x \log x)$.Using the property $\log(ab) = \l

Vedclass · Accepted Answer

$(x \log x)^{\log \log x} \left\{ \frac{1}{x \log x}(\log x + \log \log x) + (\log \log x) \left( \frac{1}{x} + \frac{1}{x \log x} \right) \right\}$

Vedclass · Answer

(A) Given $y = (x \log x)^{\log \log x}$.Taking logarithm on both sides: $\log y = \log \log x \cdot \log(x \log x)$.Using the property $\log(ab) = \log a + \log b$,we get $\log y = \log \log x (\log x + \log \log x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log \log x) \cdot (\log x + \log \log x) + \log \log x \cdot \frac{d}{dx}(\log x + \log \log x)$.$\frac{1}{y} \frac{dy}{dx} = \left( \frac{1}{\log x} \cdot \frac{1}{x} \right) (\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{\log x} \cdot \frac{1}{x} \right)$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x \log x}(\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{x \log x} \right)$.Therefore,$\frac{dy}{dx} = y \left\{ \frac{1}{x \log x}(\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{x \log x} \right) \right\}$.Substituting $y = (x \log x)^{\log \log x}$,we get the final expression.

If $y = (x \log x)^{\log \log x}$,then $\frac{dy}{dx} = $

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